The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

Answer: Option C

Explanation:

Let each side be a cm, then
$$\begin{aligned} \left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\ <=>\left(a^2-\frac{a^2}{4}\right) = 100 \\ <=> \frac{3a^2}{4} = 100 \\ a^2 = \frac{400}{3} \\ Area = \frac{\sqrt{3}}{4}*a^2 \\ = \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\ = \frac{100}{\sqrt{3}}cm^2 \end{aligned}$$

Answer: Option C

Explanation:

$$\begin{aligned} \text{Area of triangle, A1 = }\frac{1}{2}*base*height \\ = \frac{1}{2}*15*12 = 90 cm^2 \\ \text{Area of second triangle =} 2*A1 \\ = 180 cm^2 \\ \frac{1}{2}*20*height = 180 \\ => height = 18 cm \end{aligned}$$

Answer: Option B

Explanation:

In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference

So we will be having distance and time, from which we can calculate the speed. So let solve.

Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be

$$\begin{aligned} \text{40 * Circumference } \\ = \text{40 * 2*\pi*r } \\ = 40 * 2* \frac{22}{7}* 35 \\ = 8800 cm \\ = \frac{8800}{100} m = 88 m\\ \text{Distance covered in 1 sec =}\\ \frac{88}{10} \\ = 8.8 m \\ Speed = 8.8 m/s \\ = 8.8*\frac{18}{5} = 31.68 km/hr \end{aligned}$$

Answer: Option B

Explanation:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =
$$\begin{aligned} \frac{\text{Area of floor}}{\text{Area of tile}} \\ = \left(\frac{1517\times902}{41 \times 41}\right)\\ = 814 \end{aligned}$$

Answer: Option D

Explanation:

Let original length = x metres and original breadth = y metres.

$$\begin{aligned} \text{Original area } = \text{xy } m^2 \\ \text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\ \text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\ =>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\ =>\text{New Area }= \frac{36}{25}xy \\ \text{Area Difference} = \frac{36}{25}xy - xy \\ = \frac{11}{25}xy \\ Increase \% = \frac{Differnce}{Actual}*100 \\ = \frac{11xy}{25}*\frac{1}{xy}*100 = 44\% \end{aligned}$$